## University Calculus: Early Transcendentals (3rd Edition)

$-\frac{1}{10}cos 5x+\frac{cos x}{2}+C$
Applying the product-to-sum identity, we get $\int sin2x cos 3xdx=\frac{1}{2}\int[sin(-x)+sin5x]dx$ $=\frac{1}{2}\int(sin 5x- sinx)dx$ $= \frac{1}{2}\int sin 5xdx-\frac{1}{2}\int sin xdx$ $=(\frac{1}{2}\times\frac{-cos5x}{5})-(\frac{1}{2}\times-cos x)+C$ $= -\frac{1}{10}cos 5x+\frac{cos x}{2}+C$